The problem I am trying to do is the following:
Show that vector representation 5 and adjoint representation 10 in SO(5) decompose respectively into representations of SO(4) as:
5 →4⊕1
10→6⊕4
I understand that 5 is rep of SO(5) corresponding to Dynkin labels (1, 0). 1 is of course trivial representation. But what type of rep is 4? Once I started calculating weights I found out that Dynkin labels corresponding to (2, 0) as well as (1, 1) for SO(4) both have 4 weights (and different ones). And none of them seem to reproduce 5! Maybe there is an easier way, how would you show these decomposition?
My results so far:
Weights for 5 [Dynkin (1,0) for SO(5)]: (1, 0), (-1, 2), (0,0), (1, -2) (0, -1)
Weights for 4? [Dynkin (1,1) for SO(4)]: (1,1), (-1, 1), (1, -1), (-1, -1)
Weights for 4? [Dynkin (3, 0) for SO(4)]: (3, 0), (1, 0), (-1, 0), (-3, 0)
I would not do this in terms of weights but directly. You start by taking the representation $\mathbb R^5$ of $SO(5)$ and looking at the stabilizer of a unit vector. This acts trivially on the line spanned by that vector and looking at the action on the orthocomplement of that line identifies the stabilizer with $SO(4)$. This already gives you the decomposition of $\mathbb R^5\cong\mathbb R^4\oplus\mathbb R$. For the adjoint representation, you use that $\mathfrak{so}(5)\cong \Lambda^2\mathbb R^5$ as a representation of $SO(5)$. Inserting the decomposition of $\mathbb R^5$ readily leads to $\Lambda^2\mathbb R^5=\Lambda^2\mathbb R^4\oplus(\mathbb R^4\otimes\mathbb R)$, which is the direct sum of the adjoint representation and the standard representation of $SO(4)$. (This is not the complete decomposition into irreducibles, since $\mathfrak{so}(4)$ is not irreducible but splits into a direct sum of two representations of dimension $3$.)